replace long string of variables

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Cristian Martin
Cristian Martin le 1 Juin 2022
Commenté : Cristian Martin le 1 Juin 2022
Hi,
I have a script like below. The problem is that I must to create until index50, how can I write the script more easier and faster?
index1= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '1');
val1 = unique(index1);
val_unica1 = nnz(val1);
index2= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '2');
val2 = unique(index2);
val_unica2 = nnz(val2);
index3= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '3');
val3 = unique(index3);
val_unica3 = nnz(val3);
total= sum(val_unica1) + sum(val_unica2) + sum(val_unica3);
  2 commentaires
Stephen23
Stephen23 le 1 Juin 2022
Modifié(e) : Stephen23 le 1 Juin 2022
Very interesting code. Because STRCMP returns a boolean array:
index1= strcmp(a(:,1), 'mama') & strcmp(a(:,3), '1');
this line:
val1 = unique(index1);
will return one of [], FALSE, TRUE, or [FALSE,TRUE]. You then count how many times TRUE occurs:
val_unica1 = nnz(val1);
which will return either 0 or 1. It would be simpler and more efficient to use ANY.
Cristian Martin
Cristian Martin le 1 Juin 2022
Modifié(e) : Cristian Martin le 1 Juin 2022
Yes, because I want for the below matrix to set an edit text for counting a sum of unique entry for mama 1, mama2, mama3 and so on
a = {'mama', '5', '1';
'mama', '4', '1';
'mama', '6', '1';
'mama', '7', '2';
'mama', '7', '2';
'mama', '7', '1';
'mama', '7', '3';
'mama', '7', '3';
'mama', '8', '1'};
disp(total);
3

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Réponse acceptée

Steven Lord
Steven Lord le 1 Juin 2022
a = {'mama', '5', '1';
'mama', '4', '1';
'mama', '6', '1';
'mama', '7', '2';
'mama', '7', '2';
'mama', '7', '1';
'mama', '7', '3';
'mama', '7', '3';
'mama', '8', '1'};
rowsWithMama = strcmp(a(:, 1), 'mama')
rowsWithMama = 9×1 logical array
1 1 1 1 1 1 1 1 1
rowsWith1 = strcmp(a(:, 3), '1')
rowsWith1 = 9×1 logical array
1 1 1 0 0 1 0 0 1
nnz(rowsWithMama & rowsWith1)
ans = 5
If you have multiple values you want to test for in column 3 consider using ismember.
rowsWith2Or3 = ismember(a(:, 3), {'2', '3'})
rowsWith2Or3 = 9×1 logical array
0 0 0 1 1 0 1 1 0
or, using string to make creation of the list of search terms easier by avoiding needing to manually create '2', '3', '4', ... '50':
s = string(2:3)
s = 1×2 string array
"2" "3"
rowsWith2Or3 = ismember(a(:, 3), s)
rowsWith2Or3 = 9×1 logical array
0 0 0 1 1 0 1 1 0
  1 commentaire
Cristian Martin
Cristian Martin le 1 Juin 2022
Thank you for the valuable information!

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